Question: We know that $\frac{1}{1-x}=1+x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{n-1}}+...$ for $x\in\left(-1,1\right)$. Using this fact, find the function that corresponds to the following series. $ -x-\frac{{{x}^{2}}}{2}-\frac{{{x}^3}}{3}-...-\frac{{{x}^{n}}}{n}-...$ Choose 1 answer: Choose 1 answer: (Choice A) A $\frac{-1}{{{\left( 1-x \right)}^{2}}}$ (Choice B) B $\frac{1}{{{\left( 1-x \right)}^{2}}}$ (Choice C) C $ -\arctan \left( 1-x \right)$ (Choice D) D $\arctan \left( 1-x \right)$ (Choice E) E $-\ln \left( 1-x \right)$ (Choice F) F $\ln \left( 1-x \right)$
First, note that the derivative of $ -x-\frac{{{x}^{2}}}{2}-\frac{{{x}^3}}{3}-...-\frac{{{x}^{n}}}{n}-...$ is $-1-x-x^2-x^3-x^4-...=-\dfrac{1}{1-x}$ That is, $\frac{d}{dx}\left( -x-\frac{x^2}{2}-\frac{x^3}{3}-...-\frac{x^n}{n}...\right)=-\frac{1}{1-x}$ Antidifferentiate both sides. On the right-hand side, use a $~u$ -substitution $~u=1-x~$ and $~du=-dx\,$. Then $-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...-\frac{x^n}{n}-...=-\int{\frac{1}{1-x}}\text{ }dx=\ln \left( 1-x \right)+C\,$ Now let $~x=0~$ and equate the left- and right-hand sides of the last equation to see that $~C=0\,$. Thus, $-x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-...-\frac{{{x}^{n}}}{n}-...=\ln \left( 1-x \right)$